it seems like we're allowed to pick $C_1=1$ and $C_2=0$. So we pick $\psi_n(x)=\sin nx $ and $\lambda_n = n^{-2}$. Then Mercer's theorem actually says that we should get:
$$ \min(x,y)=\sum_{n=1}^\infty n^{-2}\sin nx\sin ny$$

this all seem very nice, but when evaluating this numerically, it doesn't work.
I tried also to normalize $\psi$ by dividing by its norm which is $\sqrt {\frac 1 {4n} (2nT-\sin2nT)}$, and it didn't help.

I also tried to substitute the original solution with $C_1,C_2$ in the original eigenvalue problem equation, and then to calculate $C_1,C_2$, but they turned out to depend on $x$, which is of cource unacceptable.

Similarly, from $$\lambda\psi'(x) = \int_x^T \psi(y)dy$$ you get $$\psi'(T)=0.$$

Hence, you have two boundary conditions for the differential equation $\lambda\psi''(x) = -\psi(x)$. The first forces $C_2=0$, the second gives $$\lambda = \frac{T^2}{\pi^2(n+\tfrac{1}{2})^2}$$
($T/\sqrt{\lambda}$ has to be a root of $\cos$) and no condition on $C_1$. Since you want an orthonormal basis, you have to normalize the functions in $L^2([0,T])$ which gives $C_1=\sqrt{2/T}$.

What you are missing in your numerics is that the series starts with $n=0$ and hence, your result differs from $\min(x,y)$ by $\psi_0(x)\psi_0(y) = \sin(\tfrac{\pi x}{2T})\sin(\tfrac{\pi y}{2T})$.